$x^2+y^2+z^2$ is an algebraic expression. It is given that the values of all three literals are expressed in three equations as follows.

$(1) \,\,\,\,\,\,$ $x = r\cos{\alpha}\cos{\beta}$

$(2) \,\,\,\,\,\,$ $y = r\cos{\alpha}\sin{\beta}$

$(3) \,\,\,\,\,\,$ $z = r\sin{\alpha}$

It is asked us to find the value $x^2+y^2+z^2$ in this problem on the basis of the above three equations.

Replace the algebraic expression by the respective values of the literals $x$, $y$ and $z$ to find the value of algebraic expression.

$x^2+y^2+z^2$ $\,=\,$ ${(r\cos{\alpha}\cos{\beta})}^2$ $+$ ${(r\cos{\alpha}\sin{\beta})}^2$ $+$ ${(r\sin{\alpha})}$

$=\,\,\,$ $r^2\cos^2{\alpha}\cos^2{\beta}$ $+$ $r^2\cos^2{\alpha}\sin^2{\beta}$ $+$ $r^2\sin^2{\alpha}$

Observe the first two terms in the expression, $r^2\cos^2{\alpha}$ is a common factor in both terms in the expression. So, they can be taken common from them. It is not only useful to simplify the expression and helpful to us to find the value of the expression easily.

$=\,\,\,$ $r^2\cos^2{\alpha}{(\cos^2{\beta}+\sin^2{\beta})}$ $+$ $r^2\sin^2{\alpha}$

According to Pythagorean identity of sin and cos functions, the value of $\cos^2{\beta}+\sin^2{\beta}$ is equal to one.

$=\,\,\,$ $r^2\cos^2{\alpha}{(1)}$ $+$ $r^2\sin^2{\alpha}$

$=\,\,\,$ $r^2\cos^2{\alpha}$ $+$ $r^2\sin^2{\alpha}$

This time, $r^2$ is a common factor in the two terms of the expression. So, take it common from them to proceed further in simplifying the trigonometric expression.

$=\,\,\,$ $r^2{(\cos^2{\alpha}+\sin^2{\alpha})}$

Now, use Pythagorean identity of sin and cos functions to get the complete solution of this problem.

$=\,\,\,$ $r^2{(1)}$

$=\,\,\,$ $r^2$

Therefore, it is solved that the value of $x^2+y^2+z^2$ is equal to $r^2$.

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